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Rotation problem

  1. May 27, 2007 #1
    1. The problem statement, all variables and given/known data
    A 30cm diameter wheel rolls without slipping at 120rpm. The point of contact with the ground has an instantaneous speed of:

    a-5.2mm/s
    b-1.04cm/s
    c-2.6 m/s
    d-zero


    2. Relevant equations
    I'm not really sure about my final answer but i think is d-zero because i have a diagram of a wheels that has 0m/s on the bottom of the wheel.



    [​IMG]

    [​IMG]


    3. The attempt at a solution

    If the correct answer is not zero i think i will need the following formula

    I=MR^2

    but i'm only given diameter and rpm's so i don't know if there's anohter formula i'm missing.
     
    Last edited: May 27, 2007
  2. jcsd
  3. May 27, 2007 #2

    Pythagorean

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    for each rotation the wheel goes around, how many cm of wheel touch the ground?

    I=MR^2 is inertia; you shouldn't need that for this, you're already given the speed of rotation and the geometry of the wheel

    This starts as a geometry problem.
     
  4. May 27, 2007 #3
    okay... so i might need to multiply it by 2*3.1426 or something right?
     
  5. May 27, 2007 #4

    Pythagorean

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    the equation for the perimeter of a circle is P = 2*pi*r

    where r is the radius. You're given the diameter.
     
  6. May 27, 2007 #5
     
  7. May 27, 2007 #6

    Pythagorean

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    no, 30 cm is the diameter. The diameter is 2 times the radius.

    The radius relates to the perimeter as stated in the above equation.
     
    Last edited: May 27, 2007
  8. May 27, 2007 #7

    D H

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    Newton9, your first answer was corret. "Rolling withing slipping" means the point of contact has zero instantaneous velocity. The object would be slipping if the contact point had a non-zero velocity.
     
  9. May 27, 2007 #8
    thanks guys :smile:
     
  10. May 27, 2007 #9

    Pythagorean

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    hrm, I must have a bad definition of 'point of contact', because I'd think if a wheel is rolling along, the position where it touches the floor is the point of contact, and if it's not slipping, then both the wheel and the floor are touching in different places at ever time step, and that would imply the point of contact moves.

    Can someone help me with my misconception here?
     
  11. May 27, 2007 #10

    Kurdt

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    If you look at the third diagram given in the first post you can see that the centre of the wheel has a constant speed v (translational velocity). the top of the wheel has a speed of 2v which is the tangent velocity of the point on the rim of the wheel and the translational velocity. Thus the point at the bottom of the wheel (the point of contact) has a tangental velocity that is opposite to the translational velocity and the two cancel.
     
  12. May 27, 2007 #11

    Pythagorean

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    wow... that's kind of unintuitive. It almost seems as if you're in a frame with a velocity equal to the translational velocity of the wheel.
     
  13. May 27, 2007 #12

    Kurdt

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    There's a famous photo of a wheel with its centre as a green light and a point on the rim as a red light and it illustrates it beautifully but I'm having trouble finding it on the net. If I do find one i'll give you the link and perhaps it will shed more light on it for you.

    EDIT: this helps me anyway.

    [​IMG]

    Apologies for the crude photo but I gave up on the internet.
     
    Last edited: May 27, 2007
  14. May 27, 2007 #13

    Pythagorean

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    Much appreciated, thank you!
     
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